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1. The radioisotope Cobalt-60 is used in cancer therapy. The half-life of this isotope is 5.27 years.

Which equation determines the percent of an initial amount of the isotope remaining after t years?

A=5.27(12)100tA=5.27(12)100t

A=100(12)t5.27A=100(12)t5.27

A=5.27(12)t100A=5.27(12)t100

A=100(12)5.27t

My Answer-'C'

2. What are the real and imaginary parts of the complex number?

9+7i

The real part: 

The imaginary part: 

My Answer- Real part=9, Imag. part=7i

 

3. 

What is the simplest radical form of the expression?

(x4y7)23

xy2xy−−√3xy2xy3

x6y10y√x6y10y

x2y4x2y2−−−−√3x2y4x2y23

x3y5x√

My Answer-'D'

 

Thanks so much

Guest Jan 16, 2018
 #1
avatar+89876 
+1

The correct  equation in the first part should be

 

A  =   100(1/2)t/5.27   

 

To  see that this is correct....let  t  =  5.27  and we have that

 

A  = 100(1/2) 5.27/5.27   =  100  (1/2)1  =  100 (1/2)   =  50%  left  after 5.27 years, the half-life  

 

 

Your   second answer is correct

 

I can't tell exactly what the third one is....sorry....!!!

 

 

cool cool cool

CPhill  Jan 16, 2018

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